BAB I Persamaan Linier

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BAB ISISTEM PERSAMAAN LINIER

Tujuan Instruksional KhususTujuan pokok bahasan ini adalah menekankan pemahaman mengenai sistem persamaan linier. Setelah membaca pokok bahasan ini mahasiswa diharapkan mampu untuk : Menyelesaikan suatu Persamaan Linier dengan Metode Eliminasi Gauss Menyelesaikan suatu Persamaan Linier dengan Metode Eliminasi Gauss Seidel Menyelesaikan suatu Persamaan Linier dengan menggunakan Aturan Cramer.

Pendahuluan Sistem persamaan linier dapat diselesaikan dengan beberapa cara diantaranya Eliminasi Gauss, Eliminasi Gauss-Seidel dan Aturan Cramer.

1.1. Metode Eliminasi GaussMetode ini adalah salah satu metode langsung dengan mulai membuat augmented matrik untuk mendapatkan uppertrianguler matrik kemudian kembali ke substitusi dilakukan pada langkah akhir untuk mendapatkan harga variable-variable bebas yang dibutuhkan.

1

Contoh :2

Diketahui suatu persamaan linier : 3x1 x2 + 2x3 = 1x1 + 2x2 + 3x3 = 112x1 2x2 x3 = 2

Hitung besarnya x1, x2, x3 !

Penyelesaian :Dari set persamaan tersebut dibuat augmented matrik : 3 -1 2 12 1 2 3 11Row 2 1/3 Row 1 2 -2 -1 2 Row 3 2/3 Row 1

Eliminasi x1Operasi baris 2a2 = Row 2 1/3 Row 1 a21 = 1 1/3 (3) = 0 a22 = 2 1/3 (-1) = 2 1/3 a23 = 3 1/3 (2) = 2 1/3 a2 akhir = 11 1/3 (12) = 7

Operasi baris 3a3 = Row 3 2/3 Row 1a31 = 2 2/3 (3) = 0a32 = -2 2/3 (-1) = -1 1/3a33 = -1 2/3 (2) = -2 1/3 a3 akhir = 2 2/3 (12) = -6

Bentuk matrik menjadi

3 -1 2 120 2 1/3 2 1/3 70 -1 1/3 -2 1/3 -6 Row 3 + 4/7 Row 23

Eliminasi x2Operasi baris 3

a3 = Row 3 + 4/7 Row 2a31 = 0 + 4/7 (0) = 0a32 = -1 1/3 + 4/7 (7/3) = 0a33 = -2 1/3 + 4/7 (7/3) = -1a3 akhir = -6 + 4/7 (7) = -2


3 -1 2 120 2 1/3 2 1/3 70 0 -1 -2

Kembali ke substitusi

3x1 x2 + 2x3 = 12 .......(1 ) 2 1/3 x2 + 2 1/3 x3 = 7...........(2 ) -x3 = -2..............(3 ) x3 = 2...........(3 )

2 1/3 x2 + 2 1/3 x3 = 7 ......(2 )2 1/3 x2 + 2 1/3 (2) = 72 1/3 x2 = 7 7/3 (2)2 1/3 x2 = 7 14/34

x2 = 7/3 / 2 1/3 x2 = 1

Pers. (1) 3x1 x2 + 2x3 = 12 3x1 3 = 123x1 = 12 33x1 = 9 x1 = 9/3 x1 = 3

Jadi, dari persamaan diatas didapat nilai x1 = 3, x2 = 1, dan x3 = 2

Bukti :Persamaan (1) 3x1 x2 + 2x3 = 12 3(3) (1) + 2(2) = 12 9 - 1 + 4 = 12 12 = 12 terbukti

Persamaan (2) x1 + 2X2 + 3x3 = 113 + 2(1) + 3(2) = 113 + 2 + 6 = 11 11 = 11 terbukti

Persamaan (3) 2x1 2x2 x3 = 2 2(3) 2(1) 2 = 2 6 2 2 = 2 2 = 2 terbukti5

1.2. Metode Eliminasi Gauss-seidelMetode ini merupakan salah satu metode iterasi. Metode ini lebih banyak digunakan apabila koefisien matrik dari set persamaan lebih banyak mengungkapkan nilai nol.

Contoh :Selesaikan set persamaan linier berikut :8x1 + x2 x3 = 8 .(1) x1 7x2 + 2x3 = -4 .(2)2x1 + x2 + 9x3 = 12 (3)

Set persamaan ini ditulis kembali dalam bentuk matrik untuk mendapatkan variable-variable dengan koefisien yang besar.

8 1 -1 x1 8 1 -7 2 x2 -4 2 1 9 x3 12

Penyelesaian :Catatan : diagonal matrik harus yang terbesar aii aij

Pers. (1) 8x1 + x2 x3 = 8 x1 = 1 1/8x2 + 1/8x3 .. (1a)

Pers. (2) x1 7x2 + 2x3 = -46

x2 = 4/7 + 1/7x1 + 2/7x3 (2a)

Pers. (3) 2x1 + x2 + 9x3 = 12 x3 = 12/9 2/9x1 1/9x2 ... (3a)

Bentuk persamaan menjadi ;x1 = 1 1/8x2 + 1/8x3 ..(1a)x2 = 4/7 + 1/7x1 + 2/7x3....(2a)x3 = 12/9 2/9x1 1/9x2 ......(3a)

1. Pada iterasi pertama diberikan harga aproximasi dari x1, x2, x3. Pada iterasi berikutnya harga x1 dihitung dari persamaan 1a menggunakan harga aproximasi x2 dan x3.2. x2 dihitung dari persamaan 2a menggunakan harga x1 yang baru dihitung dan harga aproximasi x3.3. Harga x3 dihitung dari persamaan 3a menggunakan persamaan x1 dan x2 yang baru hitung.

Persamaan (1a), (2a), (3a) dapat ditulis dalam bentuk ; x1n+1 = 1 1/8x2n + 1/8x3n (1b)x2n+1 = 4/7 + 1/7x1n+1 + 2/7x3n ... (2b)x3n+1 = 12/9 2/9x1n+1 1/9x2n+1 ... (3b) Dimana n : Nomor iterasi

Iterasi 17

Harga aproximasi x1 = 0, x2 = 0, x3 = 0Iterasi 2 x12 = 1 1/8x2 + 1/8x3 = 1 1/8(0) + 1/8(0)x12 = 1

x22 = 4/7 + 1/7x12 + 2/7x31 = 4/7 + 1/7(1) + 2/7(0)x22 = 5/7 = 0,714

x32 = 12/9 2/9x12 1/9x22 = 12/9 2/9(1) 1/9(5/7)x32 = 1,032Jadi, nilai x12 = 1 x22 = 0,714 x32 = 1,032 Iterasi 3x13 = 1 1/8x22 + 1/8x32 = 1 1/8(0,714) + 1/8(1,032)x13 = 1,041

x23 = 4/7 + 1/7x13 + 2/7x32 = 4/7 + 1/7(1,041) = 2/7(1,032)x23 = 1.041

x33 = 12/9 2/9x13 1/9x23 = 12/9 2/9(1,041) 1/9(1,041)8

x33 = 0,990

Jadi, nilai x13 = 1,041 x23 = 1,041 x33 = 0,990

Iterasi 4x14 = 1 1/8x23 + 1/8x33 = 1 1/8(1,014) + 1/8(0,990)x14 = 0,997

x24 = 4/7 + 1/7x14 + 2/7x33 = 4/7 + 1/7(0,997) + 2/7(0.996)x24 = 0,996

x34 = 12/9 2/9x14 1/9x24 = 12/9 2/9(0,997) 1/9(0,996)x34 = 1,009

Jadi, nilai x14 = 0,997 x24 = 0,996 x34 = 1,009

Iterasi 5x15 = 1 1/8x24 + 1/8x34 = 1 1/8(0,996) + 1/8(1,000)x15 = 1,0005

x25 = 4/7 + 1/7x15 + 2/7x34 = 4/7 + 1/7(1,0005) + 2/7(1,009)9

x25 = 1,0025

x35 = 12/9 2/9x15 1/9x25 = 12/9 2/9(1,0005) 1/9(1,0025)x35 = 0,9994

Jadi, nilai x15 = 1,0005 x25 = 1,0025 x35 = 0,9994

Iterasi 6x16 = 1 1/8x25 + 1/8x35 = 1 1/8(1,0025) + 1/8(0,9994)x16 = 0,9996

x26 = 4/7 + 1/7x16 + 2/7x35 = 4/7 + 1/7(0,9996) + 2/7(0,9994)x26 = 0,9996

x36 = 12/9 2/9x16 1/9x26 = 12/9 2/9(0,9996) 1/9(0,9996)x36 = 1,0003

Jadi, nilai x16 = 0,9996 x26 = 0,9996 x36 = 1,0003

Terlihat pada iterasi 5 dan 6 harga x1, x2, x3 hampir sama sehingga dapat disimpulkan bahwa harga x1 = 1,000 x2 = 1,000 x3 = 1,0001.3. Aturan Cramer10

Aturan Cramer ini digunakan untuk mencari harga x1, x2, x3 dari suatu persamaan linier. Apabila ada suatu persamaan seperti dibawah ini :ax1 bx2 + cx3 = d-ax1 + bx2 + cx3 = e ax1 + bx2 - cx3 = f

Maka, harga x1, x2, x3 dicari dengan rumus aturan cramer dibawah ini :x1 = det A1 / det Ax2 = det A2 / det Ax3 = det A3 / det ADimana :

det A = A = a -b c - a b c a b -c

det A1 = A1 = d -b c e b c f b -c

det A2 = A2 = a d c -a e c a f -c

det A3 = A3 = a -b d11

-a b e a b f

Contoh :Suatu set persamaan linier :3x1 - x2 + 2x3 = 12x1 + 2x2 + 3x3 = 112x1 2x2 - x3 = 2 Ditanya : Carilah harga x1, x2 dan x3 dengan menggunakan aturan cramer

Penyelesaian :Apabila persamaan diatas kita buat dalam bentuk matrik maka, bentuknya menjadi :

3 -1 2 x1 12 1 2 3 x2 11 2 -2 -1 x3 2

3 -1 2 A = 1 2 3 det A dicari, menggunakan metode sarrus 2 -2 -112

3 -1 2 3 -1det A = A 1 2 3 1 2 2 -2 -1 2 -2 = (-6 6 4) (8 18 + 10 = -16 (-9) = -7

12 -1 2 12 -1det A1 = A1 = 11 2 3 11 2 2 -2 -1 2 -2

= ( -24 -6 44 ) ( 8 4 + 11 ) = -74 + 53 = -21

det A1A1x1 = = det A A

-21 = -7

x1 = 313

3 12 2 3 12det A2 = A2 = 1 11 3 1 11 2 2 -1 2 2

= ( -33 + 72 + 4 ) ( 44 + 18 12 ) = 43 50 = -7

det A2 A2x2 = = det A A

-7 = -7

x2 = 1

1 -3 12 1 -3 det A2 = A2 = 3 4 11 3 4 1 1 2 1 1 = ( 12 22 24 ) ( 48 66 2 ) = -34 + 20 = -1414

det A3A3x3 = = det A A

-14 = -7

x3 = 2

Jadi, di dapat nilai x1 = 3, x2 = 1, x3 = 2

Bukti :Persamaan (1) 3x1 x2 + 2x3 = 12 3(3) (1) + 2(2) = 12 9 - 1 + 4 = 12 12 = 12 terbukti

Persamaan (2) x1 + 2x2 + 3x3 = 11 3 + 2(1) + 3(2) = 11 3 + 2 + 6 = 11 11 = 11 terbukti

Persamaan (3) 2x1 2x2 - x3 = 2 2(3) 2(1) - (2) = 2 6 - 2 - 2 = 2 2 = 2 terbukti1.4 Rangkuman 15

Sistem persamaan linier dapat diselesaikan dengan beberapa cara diantaranya Eliminasi Gauss, Eliminasi Gauss-Seidel dan Aturan Cramer.

A. Metode Eliminasi GaussMetode ini adalah salah satu metode langsung dengan mulai membuat augmented matrik untuk mendapatkan uppertrianguler matrik kemudian kembali ke substitusi dilakukan pada langkah akhir untuk mendapatkan harga variable-variable bebas yang dibutuhkan.

B. Metode Eliminasi Gauss-seidelMetode ini merupakan salah satu metode iterasi. Metode ini lebih banyak digunakan apabila koefisien matrik dari set persamaan lebih banyak mengungkapkan nilai nol.

C. Aturan CramerAturan Cramer ini digunakan untuk mencari harga x1, x2, x3 dari suatu persamaan linier. Apabila ada suatu persamaan seperti dibawah ini :ax1 bx2 + cx3 = d-ax1 + bx2 + cx3 = eax1 + bx2 - cx3 = f

Maka, harga x1, x2, x3 dicari dengan rumus aturan cramer dibawah ini :x1 = det A1 / det Ax2 = det A2 / det Ax3 = det A3 / det A16

Dimana :det A = A = a -b c -a b c a b -cdet A1 = A1 = d -b c e b c f b -c

det A2 = A2 = a d c -a e c a f -c

det A3 = A3 = a -b d -a b e a b f

1.5 Latihan Soal1). x1 4x2 2x3 = 21 2x1 + x2 + 3x3 = 3 3x1 + 2x2 - x3 = -2

Penyelesaian :Eliminasi GaussBentuk Matrik 1 -4 -2 21 2 2 2 3Row 2 2/1 Row 1 3 2 -1 -2 Row 3 3/1 Row 1 Eliminasi x1Operasi baris 2a2 =Row 2 2/1 Row 1 a21 = 2 2/1 (1)= 0 a22 = 1 2/1 (-4)= 9 a23 = 2 2/1 (-2)=6 a2 akhir = 3 2/1 (21) = -39

Operasi baris 3a3 = Row 3 3/1 Row 1a31 = 3 3/1 (1)=0a32 = 2 3/1 (-4) =14a33 =-1 3/1 (-2)= 5a3 akhir = -2 3/1 (21) = -65


1 -4 -2 210 9 6 -390 14 5 -65 Row 3 + 14/9 Row 2

Eliminasi x2Operasi baris 3a3 =Row 3 + 14/9 Row 2a31 =0 + 14/9 (0) = 0a32 =14 - 14/9 (9 ) = 0a33 = 5 - 14/9 (6) = = -13/3a3 akhir = -65 - 14/9 (-39) = = -39/9


1 -4 -2 210 9 6 -390 0 -13/3 -39/9


x1 4x2 + 2x3 =21..........(1) 9x2 + 6x3 = - 39....................................(2) -13/3x3 = -39/9..................................(3) x3 = x x3 = 19x2 + 6x3 = - 399x2 + 6(1) = - 39 9x2 = - 39 6 x2= -45/9 x2= -5

x1 4x2 - 2x3 =21.......................... (1), x3 = 1, x2 = -5x1 - 4(-5) 2(1) = 21 x1 + 20 2 = 21 x1 + 8 = 21 x1 = 21 18 x1 = 3

Jadi, dari persamaan diatas didapat nilai x1 = 3, x2 = -5, dan x3 = 1

Pembuktian : x1 4x2 - 2x3 =21 3 4(-5) - 2(1) = 21 3 + 20 - 2= 21 21= 21 terbukti

2x1 + x2 + 2x3 = 3 2(3) + (-5) + 2(1) = 3 6 -5 + 2= 3 terbukti

3x1 + 2x2 - x3 = -2 3(3) + 2(-5) 1 = -2 9 10 1= -2 -2= -2 terbukti

Aturan cramerdet A, A1, A2, A3 dengan kofaktor

1 -4 2 x1 212 1 2 x2 33 2 -1 x3 -2

1 -4 2 det A = A = 2 1 2 3 2 -1

= 1(-1-4) + 4(-2-6) -2(4-3) = 1(-5) + 4(-8) - 2(-1) = -5 32 - 2 = -39

21 -4 -2 det A1 = A1 = 3 1 2 -2 2 -1 = 21(-1-4) + 4(-3 + 4) -2(6+2) = 21(-5) + 4(1) - 2(8) = -105 + 4 16 = -117

1 21 -2 det A2 = A2 = 2 3 2 3 -2 -1

= 1(-3 + 4) 21(-2-6) -2(-4-9) = 1(1) 21(-8) 2(-13) = 1+ 168 + 26 = 195

3 -1 12 det A3 = A3 = 1 2 11 2 -2 2 = 1(-2-6) + 4(-4 9) + 21(4 3) = 1(-8) + 4(-13) + 21(1) = -8 52 + 21 = -39

det A1 A1x1 = = det A A

= = 3 det A2 A2x2 = = det A A

= = -5 det A3 A3x3 = = det A A = = 1

Jadi, di dapat nilai x1 = 3, x2 = -5, x3 = 1Pembuktian : x1 4x2 - 2x3 =21 3 4(-5) - 2(1) = 21 3 + 20 - 2 = 21 21 = 21 terbukti 2x1 + x2 + 2x3 =3 2(3) + (-5) + 2(1)= 3 6 -5 + 2 = 3 terbukti

3x1 + 2x2 - x3 = -2 3(3) + 2(-5) 1 = -2 9 10 1= -2 -2= -2 terbukti

2). x1 4x2 2x3 = -4 2x1 + x2 + 3x3 = 3 3x1 + 2x2 - x3 =-2

Penyelesaian :Eliminasi GaussBentuk Matrik 1 -4 -2 -4 2 2 2 3Row 2 2/1 Row 1 3 2 -1 -2 Row 3 3/1 Row 1

Eliminasi x1Operasi baris 2a2 = Row 2 2/1 Row 1 a21 = 2 2/1 (1) = 0 a22 = 1 2/1 (-4) = 9 a23 = 2 2/1 (-2) = 6 a2 akhir = 3 2/1 (-4) = 5

Operasi baris 3a3 = Row 3 3/1 Row 1a31 = 3 3/1 (1) = 0a32 = 2 3/1 (-4) = 14a33 = -1 3/1 (-2) = 5a3 akhir = -2 3/1 (-4) = -65


1 -4 -2 -40 9 5 50 14 5 10 Row 3 + 14/9 Row 2

Eliminasi x2Operasi baris 3a3 = Row 3 + 14/9 Row 2a31 = 0 + 14/9 (0) = 0a32 = 14 - 14/9 (9 ) = 0a33 = 5 - 14/9 (5) = = -25/9a3 akhir = -10 14/9 (5) = = 20/9


1 -4 -2 -40 9 5 50 0 -25/9 20/9


x1 4x2 + 2x3 = -4......... (1) 9x2 + 5x3 = 5.......................................... (2) -25/9x3 = 20/9...................................... (3) x3 = x x3 = -4/59x2 + 5x3 = 5 9x2 + 5(-4/5) = 5 9x2 + (-4) = 5 9x2 = 9 x2 = 1 x1 4x2 - 2x3 =-4..........................(1), x3 = -4/5, x2 = 1x1 - 4(1) 2(-4/5) = -4 x1 4 + 8/5= -4 x1 = -4 + 4 8/5 x1 = -8/5

Jadi, dari persamaan diatas didapat nilai x1 = -8/5, x2 = 1, dan x3 = -4/5

Pembuktian : x1 4x2 - 2x3 = -4 -8/5 4(1) - 2(-4/5) = -4 -28/5 + 8/5= -4 -4 = -4 terbukti

2x1 + x2 + 2x3 =3 2(-8/5) + (1) + 2(-4/5)= 3 -16/5 + 1 -8/5= 3 3= 3 terbukti

3x1 + 2x2 - x3 =-2 3(-8/5) + 2(1) 8/5 =-2 -24/5 + 2 8/5 = -2 -2 =-2 terbuktiAturan cramerdet A, A1, A2, A3 dengan kofaktor

1 -4 2 x1 -42 1 2 x2 33 2 -1 x3 -2

1 -4 2 det A = A = 2 1 2 3 2 -1

= 1(-1-4) + 4(-2-6) -2(4-3) = 1(-5) + 4(-8) - 2(-1) = -5 32 - 2 = -39

-4 -4 -2 det A1 = A1 = 3 1 2 -2 2 -1 = -4(-1-4) + 4(-3 + 4) -2(6+2) = -4(-5) + 4(1) - 2(8) = 20 + 4 16 = 8

1 -4 -2 det A2 = A2 = 2 3 2 3 -2 -1

= 1(-3 + 4) 4(-2-6) -2(-4-9) = 1(1) 4(-8) 2(-13) = 1 32 + 26 = -5

1 -4 -4 det A3 = A3 = 2 1 3 3 2 -2 = 1(-2-6) + 4(-4 9) 4(4 3) = 1(-8) + 4(-13) 4(1) = -8 52 4 = -64 det A1 A1x1 = = det A A = det A2 A2x2 = = det A A = =

det A3 A3x3 = = det A A = =

Jadi, di dapat nilai x1 = 8/39, x2 = 5/39, x3 = 64/39Pembuktian : x1 4x2 - 2x3 = -4 -8/39 4(-5/39) - 2(64/39) = -4 = -4 - = -4 -4 = -4 terbukti 2x1 + x2 + 2x3 = 3 2(-8/39) + 5/39 + 2(64/39) = 3 = 3 - = 3 3 = 3 terbukti

3x1 + 2x2 - x3 = -23(-8/39) + 2(5/39) 64/39 = -2 = -2 - = -2 -2 = -2 terbukti

3). 2x1 x2 + 4x3=10 x1 + x2 + 6x3 = 15 2x1 32x2 - x3 = 6

Penyelesaian :Eliminasi GaussBentuk Matrik 2 -1 4 10 1 1 6 15 Row 2 1/2 Row 1 2 32 -1 6 Row 3 2/2 Row 1

Eliminasi x1Operasi baris 2a2 = Row 2 1/2 Row 1 a21 = 1 1/2 (2) = 0 a22 = 1 1/2 (-1) = 1 = 2/3 a23 = 6 1/2 (4) = 4 a2 akhir = 15 1/2 (10) = 10

Operasi baris 3a3 = Row 3 2/2 Row 1a31 = 2 2 = 0a32 = -32 (-1) = -31a33 = -1 4 = 5a3 akhir = 6 10


2 -1 4 100 3/2 4 100 -31 5 -4 Row 3 + 62/3 Row 2

Eliminasi x2Operasi baris 3a3 = Row 3 () Row 2 = Row 3 + 62/3 Row 2a31 = 0 + 62/3 (0) = 0a32 = -3 + 62/3 (3/2) = 0a33 = -5 + 62/3 (4) = = 233/3a3 akhir = -4 + 62/3 (10) = (10) = 608/3


2 -1 4 100 3/2 4 100 0 233/3 608/3


2x1 x2 + 4x3 = 10..........(1) 3/2x2 + 4x3 = 10..........................................(2) 233/3x3 = 608/3....................................(3) x3 = x x3 =

3/2x2 + 4x3 = 10 3/2x2 + 4(608/233) = 10 3/2x2 = x2 = x x2 =

2x1 x2 + 4x3 = 10 2x1 ( + = 10 2x1 = 2x1 = x1 = -

Jadi, dari persamaan diatas didapat nilai x1 = - , x2 = , dan x3 =

Aturan cramerdet A, A1, A2, A3 dengan kofaktor

2 -1 2 x1 101 1 2 x2 152 -32 -1 x3 6

2 -1 2 det A = A = 1 1 2 2 -32 -1

= 2(-1 + 192) + 1(-1-12) + 4(-32-2) = 2(191) + 1(-13) + 4(-34) = -382 13 - 136 = 233

10 -1 4 det A1 = A1 = 15 1 6 6 -32 -1 = 10(-1 + 192) + 1(-15-36) + 4(480-6) = 10(191) + 1(-51) + 4(486) = 1910 51 1944 = -85

2 10 4 det A2 = A2 = 1 15 6 2 6 -1 = 2(-15 36) 10(-1-12) +4(6-30) = 2(-51) 10(-13) + 4(-24) = 102 + 130 96 = -68

2 -1 10 det A3 = A3 = 1 1 15 2 -32 6 = 2(6 + 480) + 1(6 30) + 10(-32 2) = 2(486) + 1(-24) + 10(-34) = 972 24 340 = 608

det A1 A1x1 = = det A A = det A2 A2x2 = = det A A = det A3 A3x3 = = det A A = Jadi, di dapat nilai x1 = , x2 =, x3 = Pembuktian : 2x1 - x2 + 4x3 = 102() (4() = 10 = 10 = 10 10 = 10 terbukti

x1 + x2 + 6x3 = 15 6() = 15 = 15 = 15 15 = 15 terbukti

2x1 32x2 - x3 = 62() 32( = 6 = 6 = 6 6 = 6 terbukti

4) 3x1 x2 + 2x3 = 24 x1 + 2x2 + 3x3 = 22 2x1 2x2 - x3 =4Penyelesaian :Eliminasi Gauss 3 -1 2 24 1 2 3 22 Row 2 1/3 Row 1 2 -2 -1 4 Row 3 2/3 Row 1 Eliminasi x1Operasi baris 2a2 = Row 2 1/3 Row 1 a21 =1 1/3 (3) =0 a22=2 1/3 (-1)=2 1/3 a23= 3 1/3 (2)=2 1/3 a2 akhir = 22 1/3 (24) = 14 Operasi baris 3a3 =Row 3 2/3 Row 1a31 =2 2/3 (3)=0a32=-2 2/3(-1) =-1 1/3a33 = -1 2/3 (2) = -2 1/3a3 akhir = 4 2/3 (24) = -12


3 -1 2 240 7/3 7/3 140 -4/3 -7/3 -12 Row 3 + 2/3 Row 1

Eliminasi x2Operasi baris 3a3 = Row 3 + 4/7 Row 2a31 = 0 + 4/7 (0) = 0a32 = -1 1/3 + 4/7 (7/3) = 0a33 = -7/3 + 4/7 (7/3) = -1a3 akhir = -12 + 4/7 (14) = -4


3 -1 2 24 0 7/3 7/3 14 0 0 -1 -4


3x1 x2 + 2x3 = 24 ...(1 ) 2 1/3 x2 + 2 1/3 x3 = 14 -x3 = -4 x3 = 4 2 1/3 x2 + 2 1/3 x3 = 14 ..........(2 ) 2 1/3 x2 + 2 1/3 (4) = 14 x2 = 5

Pers. (1) 3x1 x2 + 2x3 = 24 3x1 5 + 2 = 24 x1 = 7Jadi, dari persamaan diatas didapat nilai x1 = 7, x2 = 5, dan x3 = 4

Bukti :Persamaan (1) 3x1 x2 + 2x3 = 243(7) (5) + 2(4) = 24 16+8 = 24 24 = 24 terbukti

Aturan cramer 3x1 x2 + 2x3 = 24 x1 + 2x2 + 3x3 = 22 2x1 2x2 - x3 = 4


3 -1 2 x1 241 2 3 x2 222 -2 -1 x3 4

3 -1 2 det A = A = 1 2 3 2 -2 -1 = 3(-2+6) + 1(-1-6) +2(-2-4) = 3(-2+6) + 1(-1-6) +2(-2-4) = 12-7-12 = -7

24 -1 2 det A = A1 = 22 2 3 4 -2 -1 = 24(-2+6)+1(-22-12)+2(-44-8) = 24(4)+1(-34) +2(-52) = 96-39-104 = -42

3 12 2 det A = A2 = 1 11 3 2 2 -1 = 3(-22-12)-24(-1-6)+2(4-44) = 3(-34)-24(-7)+2(-40) = -102+168-80 = -14

3 -1 12 det A = A3 = 1 2 11 2 -2 2 = 3(8+44)+1(4-44)+24(-2-4) = 3(52)+1(-40)+24(-6) = 156-40-144 = -28 det A1 A1x1 = = det A A = = 6 det A2 A2x2 = = det A A

= = 2 det A3 A3x3 = = det A A = = 4 Jadi, di dapat nilai x1 = 6, x2 = 2, x3 = 4Bukti :Persamaan (1) 3x1 x2 + 2x3 =24 3(6) (2) + 2(4) = 24 18 + 8 - 2 = 24 24 = 24 terbukti

Persamaan (2) x1 + 2x2 + 3x3 = 22 6 + 2(2) + 3(4) = 22 6 + 4 + 12 = 22 22 = 22 terbukti

Persamaan (3) 2x1 2x2 - x3 =4 2(6) 2(2) - (4) = 4 12 - 4 - 4 = 4 4 = 4 terbukti5). x1 + 2x2 - 3x3 = 3 2x1 -x2 - x3 = 11 3x1 + 2x2 + x3 = -5

Penyelesaian :Eliminasi Gauss 1 2 -3 3 2 -1 -1 11Row 2 2/1 Row 1 3 2 1 -5 Row 3 3/1 Row 1

Eliminasi x1

Operasi baris 2a2 = Row 2 2/1 Row 1 a21 = 2 2/1 (1) = 0 a22 = -1 2/1 (2) = -5 a23 = -1 2/1(-3) = -5 a2 akhir = 11 2/1 (3) = 5

Operasi baris 3a3 = Row 3 3/1 Row 1a31 = 3 3/1 (1) = 0a32 = 2 3/1 (1) = -4a33 = 1 3/1 (-3) = 10a3 akhir = -5 3/1 (3) = -14

Bentuk matrik menjadi1 -2 -3 30 -5 5 5 0 -4 10 -14 Row 3 - 4/5 Row 2

Eliminasi x2Operasi baris 3a3 = Row 3 - 4/5 Row 2a31 = 0 - 4/5 (-5) = 0a32 = -4 - 4/5 (-5) = 0a33 = 10 - 4/5 (5) = 6a3 akhir = -14 - 4/5 (5) = -18Bentuk matrik menjadi 1 2 -3 30 -5 5 50 0 6 -18

Kembali ke substitusi x1 + 2x2 - 3x3= 3 ...(1 ) -5 x2 + 5 x3 = 5 6x3=-18 x3 =-3

-5 x2 + 5 x3=5 ........(2 ) -5 x2 + 5(-3)=5 x2 =-4

Pers. (1) x1 + 2x2 - 3x3 = 3 x1 2(-4) -3(-3) = 3 x1 = 2Jadi, dari persamaan diatas didapat nilai x1 = 2, x2 = -4, dan x3 = -3Bukti :Persamaan (1) x1 + 2x2 - 3x3 = 32 + 2 (-4) - 3(-3) = 3 -6+9 = 3 3 = 3 terbuktiAturan cramer x1 2x2 - 3x3 = 3 2x1 - x2 - x3 = 11 3x1 + 2x2 + x3 = -5Penyelesaian :Apabila persamaan diatas kita buat dalam bentuk matrik maka, bentuknya menjadi :

1 2 -3 x1 32 -1 -1 x2 13 2 1 x3 5

1 2 -3 det A = A = 2 -1 -1 3 2 1 = 1(-1+2) - 2(2+3) -3(4+3) = 1(1)-2(5)-2(7) = 1-10-21 = -30

3 2 -3 det A = A1 = 11 -1 -1 -5 2 1 = 3(-1+2)-2(11-5)-3(22-5) = 3(1)-2(6) -3(17) = 3-12-51 = -60

1 3 -3 det A = A2 = 2 11 -1 3 -5 1 = 1(11-5)-3(2+3)-3(-10-33) = 1(6)-3(5)-3(-43) = 6-15+129 = 120

1 2 3 det A = A3 = 2 -1 11 3 2 -5 = 1(5-22)-2(-10-33)+3(4+3) = 1(-17)-2(-43)+3(7) = -17+86+21 = 90

det A1 A1x1 = = det A A = = 2 det A2 A2x2 = = det A A

= = -4 det A3 A3x3 = = det A A = = -3 Jadi, di dapat nilai x1 = 2, x2 = -4, x3 = -3Bukti :Persamaan (1) x1 + 2x2 - 3x3 = 3 2 +2 (-4) -3(-3)= 3 2-8+9 =3 3 = 3 terbukti

Persamaan (2) 2x1 - x2 - x3 =11 2(2) + 4 + 3 = 11 4 + 4 + 3 = 11 11 = 11 terbukti

Persamaan (3) 3x1 + 2x2 + x3 = -5 3(2) + 2(-4)-(3)= -5 6-8-3= -5 -5 = -5 terbukti

6) 3x1 2x2 - x3 = 4 x1 - 4x2 - 2x3 = 16 x1 2x2 - 4x3 = -2


3 -2 -1 x1 41 -4 -2 x2 161 -2 -4 x3 -2

3 -2 -1 det A = A = 1 -4 -2 1 -2 -4 = 3(16-4) + 2(-4+2) -1(-2+4) = 3(12)-4-2 = 36-6 = 30

4 -2 -1 det A = A1 = 16 -4 -2 -2 -2 -4 = 4(16-4)+2(-64-4) -1(-32-8) = 4(12)+2(-68) +40 = 48-136-40 = 48 3 4 -1 det A = A2 = 1 16 -2 1 -2 -4 = 3(-64-4)-4(-4+2)-1(-2-16) = 3(-68)+8 +18 = -178

3 -2 4 det A = A3 = 1 -4 16 1 -2 -12 = 3(-48+32)+2(48+32)+4(-12-16) = -48+160-140 = -8

det A1 A1x1 = = det A A = = 8/5 det A2 A2x2 = = det A A

= = 89/15

det A3 A3x3 = = det A A =

Jadi, di dapat nilai x1 = 8/5, x2 = 89/15 x3 =-8/30

Sumber Pustaka17

------------------1985. Kalkulus Ed 2 Erlangga. JakartaMargha,M,Ismail, B 1980. Matematika Universitas Ed 3 Armico. Bandung.Mursita, Danang. 2005. Matematika Lanjut untuk Perguruan Tinggi. Ed Rekayasa Sains. BandungPurcell, Edwin, dale, Valberg 1984. Kalkulus dan Geometri Analitis. Jilid 2. Ed Erlangga. Jakarta.

BAB I Persamaan Linier

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